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Transcript
```PHY166 Fall 2005
7 – Impuls and Momentum; Conservation
(Notions useful in the description of collisions, in particular)
Newton´s
second law
F = ma
Impuls - Momentum
relation
F∆t = m∆v ≡ ∆P
( a=∆v/∆t )
Impuls
Definitions:
Momentum conservation
in isolated systems
∆t = t2−t1 etc.
( F=0 )
Change of
momentum
1 – initial state
∆P = 0
P = const
Momentum
2 - final state.
For a system of interacting objects, adding up Newton´s second laws for all objects:
F = F1+ F2 +… = m1a1+ m2a2 +…
(Sum only external forces.
Internal forces make no contribution
according to Newton´s third law F21=−F12)
F∆t = ∆P, where P = P1+ P2 +…
In isolated systems, ( F=0 )
total momentum is conserved:
P = const
Note that P is vector!
1
Collisions and recoil
A
Two limiting cases of collisions:
B
Before (1)
A
B
After (2)
• Inelastic (objects glue together after collision; Some part of the mechanical energy is lost)
PA,1+ PB,1 = P2
mAvA,1+ mBvB,1 = (mA+ mB)v2
• Elastic (Objects rebound; Mechanical energy is conserved)
PA,1+ PB,1 = PA,2+ PB,2
EA,1+ EB,1 = EA,2+ EB,2
mAvA,1+ mBvB,1 = mAvA,2+ mBvB,2
mAv2A,1+ mBv2B,1 = mAv2A,2+ mBv2B,2
Conservation of momentum for inelastic collisions and
conservation of momentum and energy for elastic collisions
allow to solve a multitude of important problems
Recoil is a phenomenon inverse to the inelastic collision:
Forces acting between two objects cause them to move in different directions
Example: The gun and the bullet (the gun recoils back and hits the shoulder)
Whereas in the inelastic collision the energy is absorbed (mechanical energy is
transformed into heat), in the recoil the energy is released (chemical energy is
transformed into mechanical energy)
2
Elastic collision in one dimension (head-on collision)
Transform
mAvA,1+ mBvB,1 = mAvA,2+ mBvB,2
mAv2A,1+ mBv2B,1 = mAv2A,2+ mBv2B,2
m A (v A,1 − v A, 2 ) = mB (vB , 2 − vB ,1 )
as
m A (v A2 ,1 − v A2 , 2 ) = mB (vB2 , 2 − vB2 ,1 )
and divide second
equation over the first
v A,1 + v A, 2 = vB ,1 + vB , 2
or
v A,1 − vB ,1 = −(vB , 2 − v A, 2 )
(the relative velocity changes the sign
after collision)
For equal masses, mA= mB, the equation above can be combined with
v A,1 + vB ,1 = vB , 2 + v A, 2
v A, 2 = vB ,1 ,
vB , 2 = v A,1
(momentum conservation)
(colliding bodies exchange velocities)
Demonstration
3
Center of mass (CM) of an extended system
Definition:
r=
1
M
∑m r
i i
i
where M = ∑ mi total mass of a system
i
Acceleration of CM:
Ma = ∑ mi a i =
i
∑F ≡ F
i
- the net force (total external force)
i
Center of mass of an extended system moves in the same way under the influence
of the net force F as the point mass M would move in this case. If we are interested
in the motion of an object as a whole we can consider only the motion of the CM.
If the net force is zero, CM moves with a constant velocity, whereas the object, in
general, rotates around its CM.
The force of gravity can be considered as applied to CM.
4
Applications, exercizes
Problem
A golf club exerts an average force of 500 N on a 0.1 kg golf ball, but the club is in
contact with the ball for only a hundredth of a second. a) What is the magnitude of
the impuls delivered by the club? b) What is the velocity acquired by the golf ball?
Solution
Given: F=500 N, ∆t=0.01 s, m=0.1 kg
To find: a) F∆t - ? b) v - ? (this is final velocity; initial velocity is zero: v1=0, v2=v)
a) F∆t = 500 N × 0.01 s = 5 N s = 5 kg m / s
a) F∆t = m∆v = m(v2-v1) = mv;
v = F∆t/m = 5 N s / 0.1 kg = 50 m/s
Check units: N s / kg = kg m / s2 × s / kg = m/s, OK
5
Problem
Four railroad cars, all with the same mass of 20000 kg sit on a track. A fifth car of
identical mass approaches them with a velocity of 15 m/s. This car collides with
the other four cars. a) what is the initial momentum of the system? b) what is the
velocity of the five coupled cars after the collision?
Solution
Given: mA=4m, mB=m, m=20000 kg, vA,1=0, vB,1=15 m/s
To find: a) P1 - ? b) v2 - ?
a) P1 = mBvB,1 = 20000 kg 15 m/s = 300000 kg m / s
b)
mAvA,1+ mBvB,1 = (mA+ mB)v2,
v2 =
mB vB ,1
m A + mB
=
vA,1 = 0
mvB ,1
1
1
= vB ,1 = 15 m/s = 3 m/s
5m
5
5
6
Problem
Calculate the fraction of the energy lost in the railroad car collision of the problem
above
Solution
2
2
(m A + mB )v22 mvB ,1 5m(vB ,1 / 5)
Elost = E1 − E2 =
−
=
−
2
2
2
2
mvB2 ,1  1  4 mvB2 ,1 4
=
= E1
1 −  =
2  5 5 2
5
mB vB2 ,1
Fraction of the energy lost:
Elost 4
= , that is, 80%
E1 5
7
Problem
Billiard ball A moving with a speed 3 m/s in the +x direction strikes an equal-mass
ball B initially at rest. The two balls are observed to move off at 45° to the x axis,
ball A above the x axis and ball B below. What are the speeds of the two ball after
the collision?
Solution
Given: mA=mB, vA,1=3 m/s, vB,1= 0, θA,1=45°, θB,1=−45°
To find: v2,A - ? v2,B - ?
Momentum of the isolated system of two balls is conserved:
P1 = P1, A + P1, B = P2 = P2, A + P2, B
Momentum conservation along the y axis:
P1, y = 0 = P2, y = m(v2, A, y + v2, B , y )
= m(v2, A sin(45°) + v2, B sin(−45°) ) =
m
( v2 , A − v2 , B )
2
v 2 , A = v2 , B
Momentum conservation along the x axis:
P1, x ≡ mv1, A = P2, x = m(v2, A, x + v2, B , x )
= m(v2, A cos(45°) + v2, B cos(−45°) ) =
m
(v2, A + v2, B ) = 2mv2, A
2
v2 , A = v2 , B =
8
= 2.1 m/s
v1, A
2
Homework:
Giancoli Chapter 7, Problems 7, 17, 23, 29, 51
9
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